Redox

Definition: Redox Reaction

Electron transfer reactions where reduction and oxidation occur simultaneously

Definition: Reduction

• Gain of electrons, gain of hydrogen, loss of oxygen and decrease in oxidation number of an atom

  • Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s)

    • oxidation → oxidation of Zn increases from 0 to +2

Definition: Oxidation

• Loss of electrons, loss of hydrogen, gain of oxygen and increase in oxidation number of an atom.

  • Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s)

    • reduction → oxidation of Cu decreases from +2 to 0

Definition: Reducing Agent

Substance that loses one or more electrons, undergoes oxidation, and has an increase in oxidation number. It causes other atoms to undergo reduction.

Definition: Oxidising Agent

Substance that gains one or more electrons, undergoes reduction, and has a decrease in oxidation number. It causes other atoms to undergo oxidation.

Definition: Oxidation Number

Number of electrons which the atom loses / gains or tends to lose / gain when it forms a substance

• Positive → tends to lose

• Negative → tends to gain

Memorise: Format for answering Redox Reaction

• The oxidation state of Br increases from -1 in KBr to 0 in Br₂. Hence, KBr is oxidised. The oxidation state of Cl decreases from 0 in Cl₂ to -1 in KCl. Hence, Cl₂ is reduced. Since reduction and oxidation occur simultaneously, the reaction is a redox reaction.

The oxidation number of an atom in an element is always ___.

Na, Mg, Fe, P₄, O₂

zero

The oxidation number of a monatomic ion is equal to ___

the charge on the ion

Ba²⁺ = +2

Group 1 elements’ oxidation number is always ___

+1

Group 2 elements’ oxidation number is always ___

+2

Group 13 elements’ oxidation number is always ___

+3

F oxidation number is always ___

-1

H oxidation number is ___ EXCEPT in ….

+1

-1 in metal hydrides, NaH, CaH₂

O oxidation number is ___ EXCEPT in ….

-2

-1 in peroxides

+2 in OF₂

In a neutral compound, the sum of the oxidation numbers are ___

zero

Al2O3 +3, –2

2(+3) + 3(–2) = 0

In a polyatomic ion, the sum of the oxidation numbers is equal to the ____.

charge

Cr2O72- +6, –2

2(+6) + 7(–2) = –2

Definition: Electronegativity

A measure of an atom’s ability to attract electrons in a covalent bond.

• Higher electronegativity = Greater attraction for electrons

Higher electronegativity tends to …. shared electrons.

attract

Electronegativity increases … and decreases …

It increases across the period and decreases down the group.

• Fluorine = Most electronegative

• Oxygen = Second Most electronegative

Memorise: Oxidation States of Ions

Definition: Disproportionation Reaction

Redox reaction where atoms of an element in a single substance undergo reduction and oxidation simultaneously.

Memorise: Format for answering Disproportionation Reaction

• The oxidation state of Cl increases from 0 in Cl₂ to +1 in ClO^-. Hence, Cl₂ is oxidised. The oxidation state of Cl decreases from 0 in Cl₂ to -1 in Cl^-. Hence, Cl₂ is reduced. Since Cl₂ is both oxidised and reduced simultaneously, the reaction is a disproportionation reaction.

Memorise: Balancing / Forming Redox Equations

If it is in an acidic medium, use H⁺ ions to balance hydrogen and H₂O to balance oxygen. If it is in a basic medium, after using H⁺ ions to balance hydrogen, use OH^- ions to balance hydrogen and H₂O to balance oxygen.

Use the acronym (Apple of Her Eye)

• A – Balance all elements other than hydrogen and oxygen

• O – Balance oxygen by adding H₂O

• H – Balance hydrogen by adding H⁺ (for acidic) / OH^- (for basic)

• E – Balance electrons

Example: Ion Electron Half-Equation

Example 1

Cr₂O₇^2- + I^- → Cr³⁺ + I₂ (reaction done in acidic medium)
+6          –1      +3      0 

1. Write half equations containing reactants and products of the species that have undergone oxidation and reduction.

   1. Cr₂O₇^2- → Cr³⁺ [reduction]

   2. I^- → I₂ [oxidation]

2. Balance the equations for the elements that have undergone a change in oxidation state.

   1. Cr₂O₇^2- → 2 Cr³⁺

   2. 2 I^- → I₂

3. Balance the equations for oxygen by adding H₂O.

   1. Cr₂O₇^2- → 2 Cr³⁺ + 7 H₂O

   2. 2 I^- → I₂

4. Balance the equations for hydrogen by adding H⁺.

   1. 14 H⁺ + Cr₂O₇^2- → 2 Cr³⁺ + 7 H₂O

   2. 2 I^- → I₂

5. Balance the equations for the charge by adding electrons.

   1. 14 H⁺ + Cr₂O₇^2- + 6e → 2 Cr³⁺ + 7 H₂O [reduction = gain electrons / add e to left side]

   2. 2 I^- → I₂ + 2e [oxidation = lose electrons / add e to right side]

6. Multiply with the appropriate values such that the number of electrons for both equations are equal.

   1. 14 H⁺ + Cr₂O₇^2- + 6e → 2 Cr³⁺ + 7 H₂O

   2. 2 I^- → I₂ + 2e (x3)
      = 6 I^- → 3 I₂ + 6e

7. Combine to eliminate the electrons. (i.e. total electrons lost due to oxidation = total electrons gained for reduction). The final equation can now be obtained.

   1. 14 H⁺ + Cr₂O₇^2- + 6 I^- → 2 Cr³⁺ + 7 H₂O + 3 I₂

Example 2

MnO₄^- + SO₃^2- → MnO₂ + SO₄^2- (reaction done in aqueous alkaline medium)
+7           +4          +4          +6

1. Similar steps with acidic medium (1 – 6)

   1. 3e + 4 H⁺ + MnO₄^- → MnO₂ + 2 H₂O [reduction]

   2. SO₃^2- + H₂O → SO₄^2- + H⁺ + e (x3)
      = 3 SO₃^2- + 3 H₂O → 3 SO₄^2- + 3 H⁺ + 3e 

   3. 4 H⁺ + MnO₄^- + 3 SO₃^2- + 3 H₂O → MnO₂ + 2 H₂O + 3 SO₄^2- + 3 H⁺ 

2. Add OH^- to both sides to neutralize any H⁺ present.

   1. 4 OH^- + 4 H⁺ + MnO₄^- + 3 SO₃^2- + 3 H₂O → MnO₂ + 2 H₂O + 3 SO₄^2- + 3 H⁺ + 4 OH^-
      = 4 H₂O + MnO₄^- + 3 SO₃^2- + 3 H₂O → MnO₂ + 2 H₂O + 3 SO₄^2- + OH^- + 3 H₂O

3. Cancel out the extra H₂O by adding / subtracting to obtain the final equation.

   1. 4 H₂O + MnO₄^- + 3 SO₃^2- + 3 H₂O → MnO₂ + 2 H₂O + 3 SO₄^2- + OH^- + 3 H₂O
      = 7 H₂O + MnO₄^- + 3 SO₃^2- → MnO₂ + 3 SO₄^2- + OH^- + 5 H₂O
      = 2 H₂O + MnO₄^- + 3 SO₃^2- → MnO₂ + 3 SO₄^2- + OH^-

Example: Change in Oxidation Number

Example 1

Cr₂O₇^2- + SO₃^2- → Cr³⁺ + SO₄^2- (reaction done in acidic medium)
+6          +4           +3         +6

1. Identify the atoms that have undergone a change in oxidation number.

2. Find the change in oxidation number, and hence the number of moles of electrons transferred for each reactant.

   1. For Cr₂O₇^2- to Cr³⁺, there is a reduction of Cr from +6 to +3, which means 3 electrons were transferred. However, since the compound has Cr₂, to find electrons transferred per mole, we have to calculate 3 x 2, meaning 6 electrons transferred per mole.

   2. For SO₃^2- to SO₄^2-, there is an oxidation of S from +4 to +6, which means 2 electrons transferred per mole.

3. Balance the atoms that have undergone redox change. Cross multiply the 2 numbers so that total number of electrons gained = total number of electrons lost, and electrons are balanced.

   1. 
      

	Cr2O72-	SO32-
Total number of moles of electrons transferred	6 x 2 = 12 (6 Cr)	2 x 6 = 12 (2 S)

4. Use the moles of electrons transferred on the left side, and balance the other side normally.

   1. 2 Cr₂O₇^2- + 6 SO₃^2- → 4 Cr³⁺ + 6 SO₄^2-

5. Balance the rest of the atoms not involved in redox. Then, balance the charge by adding H⁺ (acidic), and balance oxygen and hydrogen by adding H₂O.

   1. 16 H⁺ + 2 Cr₂O₇^2- + 6 SO₃^2- → 4 Cr³⁺ + 6 SO₄^2- + 8 H₂O

Example 2

MnO₄^- + NO₂^- → MnO₂ + NO₃^- (reaction done in aqueous alkaline medium)
+7          +3           +4         +5

1. Similar steps with acidic medium (1 – 4)

   1. 2 MnO₄^- + 3 NO₂^- → 2 MnO₂ + 3 NO₃^-

2. Balance the rest of the atoms not involved in redox. Then, balance the charge by adding OH^- (basic), and balance oxygen and hydrogen by adding H₂O.

   1. H₂O + 2 MnO₄^- + 3 NO₂^- → 2 MnO₂ + 3 NO₃^- + 2 OH^-

Example: Deducing Unknown Oxidation States

A metallic ion, Y^m+ is oxidised to YO₄^- by Cr₂O₇^2- in acidic medium. If 3.50 × 10^-3 moles of Y^m+ require 1.00 × 10^-3 moles of Cr₂O₇^2- for complete oxidation, determine m, and construct the balanced equation for the reaction.

1. Write down the known half equation.

   1. 14 H⁺ + Cr₂O₇^2- + 6e → 2 Cr³⁺ + 7 H₂O

   2. Y’s oxidation number in YO₄^- would be +7.

2. Determine the number of moles of the two reactants involved in the redox reaction.

   1. Amount of Y^m+ = 3.50 x 10^-3 mol

   2. Amount of Cr₂O₇^2- = 1.00 x 10^-3 mol

3. Write down the mole ratio between the 2 reactants. Determine the number of moles of electrons gained (or lost) by 1 mol of the other reactant, and hence, determine the original or final oxidation number.

   1. 1 mol of Cr₂O₇^2- gains 6e while 1 mol of Y^m+ loses (7 – m) e.

   2. Since, total electrons lost = total electrons gained;
      3.50 x 10^-3 x (7 – m) = 1.00 x 10^-3 x 6
      7 - m = 1.7143
      - m = - 5.2857
      m = + 5.2857
      m ~ +5

4. Write down the final equation.

   1. Cr₂O₇^2- + 3 Y⁵⁺ + 14 H⁺ → 2 Cr³⁺ + 3 YO₄^- + 7 H₂O

Memorise: Common Oxidising Agents and What They Reduce to:

MnO4- Manganate (VII) Purple	Mn2+  Manganese (II) [acidic] Pale Pink / Colourless
	MnO2 Manganese dioxide [alkaline] Brown PPT
Cr2O72- Dichromate (VI) Orange	Cr3+ Chromium (III) Green
IO3- Iodate (V) Colourless	I2 (aq) Iodine Brown
I2 (aq) Iodine Brown	I-  Iodide Colourless
Fe3+  Iron (III) Pale Yellow	Fe2+ Iron Pale Green
H2O2  Hydrogen peroxide Colourless	H2O Water Vapour Colourless
NO2- Nitrate (III) or Nitrite Colourless	NO (g) Nitric Oxide Colourless

Memorise: Common Reducing Agents and What They Oxidise to:

Reducing Agents	Oxidised to
C2O42- Ethanedioate Colourless	CO2 Carbon dioxide
S2O32-  Thiosulfate Colourless	S4O62- Tetrathionate
H2O2 Hydrogen Peroxide Colourless	O2
NO2- Nitrite Colourless	NO3- Nitrate (V)
Cl- Colourless	Cl2 Pale Yellow
Br- Colourless	Br2 Orange
I- Colourless	I2 (aq) Brown
Fe2+ Pale Green	Fe3+ Pale Yellow

#H₂O₂ and NO₂^- are (oxidising/reducing) agents.

BOTH OXIDISING AND REDUCING

If it is on the left, it is an oxidising agent but if it is on the right, it is a reducing agent. MnO₄^- becomes Mn²⁺ in acidic, and MnO₂ in basic solutions. H₂O₂ becomes H₂O in acidic, and O₂ in basic.

Potassium Manganate (VII) Titration:

The KMnO₄ is placed in the burette and added to the reducing agent in the conical flask. MnO₄^- titrations are carried out in ______ and ______ is added in excess to the reducing agent in the conical flask. HCl and HNO₃ cannot be used because ______________, and also because _______. Common reducing agents that react with MnO₄^- are Fe²⁺, H₂O₂, C₂O₄^2-.

acidic medium

1 mol dm^-3 sulfuric acid

the Cl^- from HCl can be oxidised by MnO₄^-

HNO₃ itself is an oxidizing agent

Iodometric / Iodine-Thiosulfate Titration

Iodometric titration is used to determine the concentration of iodine or substances which liberate iodine from potassium iodide, KI, such as potassium iodate (V), KIO₃. The amount of iodine liberated can be determined by titrating with thiosulfate, S₂O₃^2-, as iodine will always react with thiosulfate.

The Oxidation and Reduction Equations are: _____________

There are two steps to an iodometric titration.

1. An oxidising agent is normally added to excess acidified KI to liberate I₂. [The colour, _____, is usually observed]

2. The liberated iodine is then estimated by titration using a standard thiosulfate, S₂O₃^2-.

The brown colour of iodine fades to _____. A _______ is added at this stage to give a ______ coloration due to formation of starch-iodine complex. Addition of ______ drop-by-drop continues until the blue black colour disappears. A _______ solution remains.

Oxidation: 2 S₂O₃^2- → S₄O₆^2- + 2e
Reduction: I₂ + 2e → 2I^-

Brown

pale yellow

starch indicator

blue black

thiosulfate

colourless